Question

Let $\lambda \in R,\vec{a}=\lambda \hat{i}+2\hat{j}-3\hat{k},\vec{b}=\hat{i}-\lambda \hat{j}+2\hat{k}$. If $((\vec{a}+\vec{b})\times (\vec{a}\times \vec{b}))\times (\vec{a}-\vec{b})=8\hat{i}-40\hat{j}-24\hat{k}$, then $|\lambda (\vec{a}+\vec{b})\times (\vec{a}-\vec{b}){|}^2$ is equal to

• A. 136
• B. 132
• C. 140
• D. 144

Answer 1

Answer: (C)

$\vec{a}=\lambda \hat{i}+2\hat{j}-3\hat{k}$
$\vec{b}=\hat{i}-\lambda \hat{j}+2\hat{k}$
$\Rightarrow (\vec{b}-\vec{a})\times ((\vec{a}+\vec{b})\times (\vec{a}\times \vec{b}))=8\hat{i}-40\hat{j}-24\hat{k}$
$\Rightarrow ((\vec{a}-\vec{b})\cdot (\vec{a}+\vec{b}))(\vec{a}\times \vec{b})=8\hat{i}-40j-24\hat{k}$
$\Rightarrow 8(\vec{a}\times \vec{b})=8\hat{i}-40\hat{j}-24\hat{k}$
Now, $\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \lambda & 2& -3\\ 1& -\lambda & 2\end{array}\right|$
$=(4-3\lambda )\hat{i}-(2\lambda +3)\hat{j}+\left(-{\lambda }^2-2\right)\hat{k}$
$\Rightarrow \lambda =1$
$\therefore \vec{a}=\hat{i}+2\hat{j}-3\hat{k}$
$\vec{b}=\hat{i}-\hat{j}+2\hat{k}$
$\Rightarrow \vec{a}+\vec{b}=2\hat{i}+\hat{j}-\hat{k},\vec{a}-\vec{b}=3\hat{j}-5\hat{k}$
$\Rightarrow (\vec{a}+\vec{b})\times (\vec{a}-\vec{b})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -1\\ 0& 3& -5\end{array}\right|=2\hat{i}+10\hat{j}+6\hat{k}$
$\therefore$ required answer $=4+100+36=140$